If radius of the solid sphere is doubled by keeping its mass constant, the ratio of their moment of inertia about any of its diameter is

Question

If radius of the solid sphere is doubled by keeping its mass constant, the ratio of their moment of inertia about any of its diameter is (A) 1: 8 (B) 2: 5 (C) 2: 3 (D) 1: 4

Tardigrade Admin · Accepted Answer

Key Idea Moment of inertia of a solid sphere about its diameter is given as,

I = \frac{2}{5} M R^{2},

where

M

and

R

are the mass and radius of the sphere, respectively.
Moment of inertia of a solid sphere about its diameter is

I = \frac{2}{5} M R^{2}

...(i)
As the radius of a solid sphere is doubled then the new moment of inertia of the sphere will be

l^{'} = \frac{2}{5} M (2 R)^{2} = \frac{8}{5} M R^{2}

...(ii)
From Eqs. (i) and (ii), we get

\frac{1}{l ^{'}} = \frac{\frac{2}{5} M R ^{2}}{\frac{8}{5} M R ^{2}} = \frac{1}{4} = 1 : 4

The ratio of their moment of inertia about any of its diameter is

1 : 4

Q. If radius of the solid sphere is doubled by keeping its mass constant, the ratio of their moment of inertia about any of its diameter is

$1 : 8$

$2 : 5$

$2 : 3$

$1 : 4$

Q. If radius of the solid sphere is doubled by keeping its mass constant, the ratio of their moment of inertia about any of its diameter is

1:8

2:5

2:3

1:4

$1 : 8$

$2 : 5$

$2 : 3$

$1 : 4$